Beautiful Proofs of Some Simple Math Propositions

Правка en2, от Andycode3759, 2024-01-13 14:54:31

I came up with them all when I was in a daze during classes in school.

This post is a practice translation for helping myself get familiar to English math terms. So if I made any mistakes, it will be grateful if you could correct them in the comments.

Proposition 1

Proposition: Convex n-polygon $$$(n \geq 4)$$$ has $$${{n(n-3)} \over 2}$$$ diagonals.

Proof: Draw a convex n-polygon in a plane randomly and we can get its $$$n$$$ vertices, among which we can connect any two of them to get $$$C_{n}^{2}$$$ line segments in total, in which $$$n$$$ of them are the edges of the polygon, and the rest $$$C_{n}^{2}-n={{n(n-3)} \over 2}$$$ of them are the diagonals. Proved in only one sentence!

Proposition 2

Proposition: There are two finite discrete random variables $$$X$$$ and $$$Y$$$ whose expectations are $$$E(X)$$$ and $$$E(Y)$$$ and variances are $$$D(X)$$$ and $$$D(Y)$$$. Then $$$E(X+Y)=E(X)+E(Y), D(X+Y)=D(X)+D(Y)$$$.

Proof:

(i) Assume the distributions of $$$X$$$ and $$$Y$$$ are $$$P(X=x_i)=p_i(1 \leq i \leq n),P(Y=y_i)=q_i(1 \leq i \leq m)$$$, then

$$$ E(X) = \sum_{i=1}^{n}p_ix_i, ~ E(Y) = \sum_{i=1}^{m}q_iy_i $$$

Obviously we have

$$$ \sum_{i=1}^{n}p_i=1, ~ \sum_{i=1}^{m}q_i=1, $$$

So

$$$ \begin{align*} E(X+Y) &= \sum_{i=1}^{n} \sum_{j=1}^{m}p_iq_j(x_i+y_j) \\ &= \sum_{i=1}^{n} p_i(x_i\sum_{j=1}^{m}q_j + \sum_{j=1}^{m}q_jy_j) \\ &= \sum_{i=1}^{n} p_i(x_i + E(Y)) \\ &= \sum_{i=1}^{n} p_ix_i + E(Y) \\ &= E(X) + E(Y) \end{align*} $$$

(ii) For convenience, take $$$dx_i$$$ as $$$x_i-E(X)$$$ and $$$dy_i$$$ as $$$y_i-E(Y)$$$. First we need a lemma:

$$$ \sum_{i=1}^{n}p_idx_i=0 $$$

(Also apply to $$$Y$$$.) The proof is simple:

$$$ \begin{align*} \sum_{i=1}^{n}p_idx_i &= \sum_{i=1}^{n}p_i(x_i-E(X)) \\ &= \sum_{i=1}^{n}p_ix_i - E(X)\sum_{i=1}^{n}p_i \\ &= E(X) - E(X) \\ &= 0 \end{align*} $$$

Since we have

$$$ D(X) = \sum_{i=1}^{n}p_i(x_i-E(X))^2 = \sum_{i=1}^{n}p_idx_i^2 \\ D(Y) = \sum_{i=1}^{n}q_i(y_i-E(Y))^2 = \sum_{i=1}^{m}q_idy_i^2 $$$

Then

$$$ \begin{align*} D(X+Y) &= \sum_{i=1}^{n} \sum_{j=1}^{m}p_iq_j(x_i+y_j-E(X+Y))^2 \\ &= \sum_{i=1}^{n} \sum_{j=1}^{m}p_iq_j(dx_i+dy_j)^2 \\ &= \sum_{i=1}^{n} p_i\sum_{j=1}^{m}(q_jdx_i^2 + 2q_jdx_idy_j + q_jdy_j^2) \\ &= \sum_{i=1}^{n} p_i (dx_i^2\sum_{j=1}^{m}q_j + 2dx_i\sum_{j=1}^{m}q_jdy_j + \sum_{j=1}^{m}q_jdy_j^2) \\ &= \sum_{i=1}^{n} p_i (dx_i^2 + 0 + D(Y)) \\ &= \sum_{i=1}^{n} p_idx_i^2 + D(Y) \\ &= D(X) + D(Y) \end{align*} $$$

Q.E.D.

Proposition 3

Proposition: $$$\forall n \in \mathbb{N}^{+}, 1^2+2^2+3^2+ \dots +n^2 = \sum_{k=1}^{n}k^2 = {1 \over 6}n(n+1)(2n+1)$$$ (The sum of squares of the first $$$n$$$ positive integers).

Proof: It is noticed that:

$$$ \begin{align*} 1^2 &= 1 \\ 2^2 &= 1+3 \\ 3^2 &= 1+3+5 \\ \dots \\ n^2 &= 1+3+5+ \dots +(2n-1) \end{align*} $$$

We can write $$$n$$$ equations like these and add them up, then the left side is $$$1^2+2^2+\dots+n^2$$$, and the right side is the sum of $$$n$$$ times $$$1$$$, $$$(n-1)$$$ times $$$3$$$, ..., $$$(n-k+1)$$$ times $$$(2k-1)$$$, ..., $$$1$$$ times $$$(2n-1)$$$. So we will get

$$$ \sum_{k=1}^{n}k^2 = \sum_{k=1}^{n}(n-k+1)(2k-1) $$$

Reduce the right side of the equation:

$$$ \begin{align*} \sum_{k=1}^{n}(n-k+1)(2k-1) &= \sum_{k=1}^{n}(2nk-2k^2+3k-n-1) \\ &= \sum_{k=1}^{n}(2nk-2k^2+3k)-n(n+1) \\ &= (2n+3)\sum_{k=1}^{n}k - 2\sum_{k=1}^{n}k^2 - n(n+1) \\ &= {n(n+1)(2n+3) \over 2} - n(n+1) - 2\sum_{k=1}^{n}k^2 \\ &= n(n+1)({2n+3 \over 2}-1) - 2\sum_{k=1}^{n}k^2 \\ &= {n(n+1)(2n+1) \over 2} - 2\sum_{k=1}^{n}k^2 \end{align*} $$$

Do the transposition:

$$$ \begin{align*} 3 \sum_{k=1}^{n}k^2 &= {n(n+1)(2n+1) \over 2} \\ \sum_{k=1}^{n}k^2 &= {n(n+1)(2n+1) \over 6} \end{align*} $$$

Q.E.D.

Теги math, proofs

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  Rev. Язык Кто Когда Δ Комментарий
en4 Английский Andycode3759 2024-01-13 17:57:34 65 a smol mistake
en3 Английский Andycode3759 2024-01-13 17:00:03 88 add original link
en2 Английский Andycode3759 2024-01-13 14:54:31 0 (published)
en1 Английский Andycode3759 2024-01-13 14:53:10 4000 Initial revision (saved to drafts)